Closed form of the recurrence relation: $x_{n+1} = {x_n \over b + x_n}$, $x_1 = a$

Question

I'm solving the following recurrence:

Find the closed form of the recurrence relation
$$
\begin{cases}
x_{n+1} = {x_n \over b + x_n} \\
x_1 = a
\end{cases}
$$
Where $a, b$ are some constants.

I've started with a substitution:

$$
z_n = {1 \over x_n}
$$

Rewrite to obtain the form of characteristic equation:
$$
z_{n+1} = bz_n + 1
$$

Solve for homogenous and particular parts: $z_n = z_n^{(h)} + z_n^{(p)}$

$$
z_n^{(h)} = C\cdot b^n \\
z_n^{(p)} = {1\over 1 -b}
$$

Particular is obtained by assuming that particular solution is some constant $B$. Now using the initial conditions $x_1 = a$ and doing reverse substitution:

$$
\begin{align}
a &= \frac{1-b}{C\cdot b^n(1-b) + 1} \iff \\
\iff C &= \frac{1-b-a}{ab(1-b)}
\end{align}
$$

Which gives the closed form for $x_n$:

$$
x_n = \frac{(1-b)^2ab}{(1-b)((1-b-a)b^n+ab)}
$$

So here if we assume that $b \ne 1$ then $x_n$:

$$
\begin{cases}
x_n = \frac{a(b-1)}{(a+b-1)b^{n-1} – a} \\
b \ne 1 \\
a \ne \frac{-(b-1)b^k}{b^k – 1} \\
k, n \in \mathbb N \\
n \ge 2
\end{cases}
$$

This matches the answer from the keys section. However keys suggest yet another solution assuming $b = 1$ and $a \ne -{1\over k}$. That solution is as follows:

$$
\begin{cases}
x_n = \frac{a}{1+(n-1)a} \\
b = 1 \\
a \ne -{1 \over k} \\
k, n \in \mathbb N
\end{cases}
$$

I don't really understand where the second solution comes from. How can we assume that $b = 1$? Won't it give a nonsense implying division by zero in the general term for $x_n$?

Answer 1

$b=1$ gives a perfectly good recurrence relation $x_{n+1}=\dfrac{x_n}{1+x_n}$ for almost all $a$. You assumed $b\neq 1$ in order to impose the guess $z_n^{(p)}=\frac{1}{b-1}$.

Indeed, the recurrence relation for $z_n$ becomes very easy to solve when $b=1$: it is just adding $1$ at every step so $z_n=z_1+n-1$.